5.6: Inhomogeneous Equations
( \newcommand{\kernel}{\mathrm{null}\,}\)
Consider a rod of length 2m, laterally insulated (heat only flows inside the rod). Initially the temperature u is
1ksin(πx2)+500 K.
The left and right ends are both attached to a thermostat, and the temperature at the left side is fixed at a temperature of 500 K and the right end at 100 K. There is also a heater attached to the rod that adds a constant heat of sin(πx2) to the rod. The differential equation describing this is inhomogeneous
∂∂tu=k∂2∂x2u+sin(πx2),u(0,t)=500,u(2,t)=100,u(x,0)=1ksin(πx2)+500.
ince the inhomogeneity is time-independent we write
u(x,t) = v(x,t) + h(x), \nonumber
where h will be determined so as to make v satisfy a homogeneous equation. Substituting this form, we find
\dfrac{\partial}{\partial t} v = k \dfrac{\partial^2}{\partial x^2} v + k h'' + \sin\left( \frac{\pi x}{2} \right). \nonumber
To make the equation for v homogeneous we require h''(x) = - \frac{1}{k}\sin\left( \frac{\pi x}{2} \right), \nonumber
which has the solution
h(x) = C_1 x + C_2 + \frac{4}{k\pi^2}\sin\left( \frac{\pi x}{2} \right). \nonumber
At the same time we let h carry the boundary conditions, h(0)=500, h(2)=100, and thus h(x) = -{200} x + 500 + \frac{4}{k\pi^2}\sin\left( \frac{\pi x}{2} \right). \nonumber The function v satisfies
\begin{aligned} \dfrac{\partial}{\partial t} v &= k \dfrac{\partial^2}{\partial x^2} v, \nonumber\\[4pt] v(0,t) & = v(\pi,t) = 0, \nonumber\\[4pt] v(x,0) & = u(x,0) - h(x) = {200} x.\end{aligned} \nonumber
This is a problem of a type that we have seen before. By separation of variables we find
v(x,t) = \sum_{n=1}^\infty b_n \exp (- \frac{n^2\pi^2}{4}kt) \sin \frac{n\pi}{2}x. \nonumber
The initial condition gives
\sum_{n=1}^\infty b_n \sin nx = {200} x. \nonumber
from which we find
b_n = (-1)^{n+1} \frac{800}{n\pi}. \nonumber
And thus
u(x,t) = -\frac{200} x + 500 + \frac{4}{\pi^2 k} \sin \left(\frac{\pi x}{2}\right) + \frac{800}{\pi} \sum_{n=1}^\infty \frac{(-1)^n}{n+1} \sin \left(\frac{\pi nx}{2}\right) e^{-k (n \pi/2)^2 t}. \label{eq.10}
Note: as t\rightarrow \infty, u(x,t) \rightarrow -\frac{400}{\pi} x + 500 + \frac{\sin \frac{\pi}{2}x}{k}. As can be seen in Fig. \PageIndex{1} this approach is quite rapid – we have chosen k=1/500 in that figure, and summed over the first 60 solutions.
Figure \PageIndex{1}: Time dependence of the solution to the inhomogeneous Equation \ref{eq.10}.